# For f(x) = 2x^2 - 4x -1, what is the vertex and axis of symmetry?

Dec 9, 2017

vertex at $\left(1 , - 3\right)$, the axis of symmetry is at $x = 1$

#### Explanation:

There are two ways of doing this - completing the square or through differentiation. Either way, we are trying to find the minimum point.

Completing the square

$f \left(x\right) = 2 {x}^{2} - 4 x - 1$
$= 2 \left({x}^{2} - 2 x - \frac{1}{2}\right)$
$= 2 \left({\left[x - 1\right]}^{2} - 1 - \frac{1}{2}\right)$
$= 2 \left({\left[x - 1\right]}^{2} - \frac{3}{2}\right)$
$= 2 {\left[x - 1\right]}^{2} - 3$

This means the minimum value is at $f \left(x\right) = - 3$, where $x = 1$.

The axis of symmetry will be the same as the x-coordinate of the minimum, this being $x = 1$

So the vertex is at $\left(1 , - 3\right)$, the axis of symmetry is at $x = 1$

Differentiation

I'd generally only differentiate if the number were a bit awkward, if completing the square would give loads of fractions.

$f \left(x\right) = 2 {x}^{2} - 4 x - 1$
$f ' \left(x\right) = 4 x - 4$
Let $f ' \left(x\right) = 0$
$4 x - 4 = 0$
$x = 1$
Let $x = 1$
$f \left(1\right) = 2 {\left(1\right)}^{2} - 4 \left(1\right) - 1$
$= - 3$

So the vertex is at $\left(1 , - 3\right)$, the axis of symmetry is at $x = 1$